Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int OO = 1e9;
const long long INF = 1e18;
const int irand(int lo, int hi) {
  return ((double)rand() / (RAND_MAX + 1.0)) * (hi - lo + 1) + lo;
}
const long long lrand(long long lo, long long hi) {
  return ((double)rand() / (RAND_MAX + 1.0)) * (hi - lo + 1) + lo;
}
template <typename T>
T getnum() {
  int sign = 1;
  T ret = 0;
  char c;
  do {
    c = getchar();
  } while (c == ' ' || c == '\n');
  if (c == '-')
    sign = -1;
  else
    ret = c - '0';
  while (1) {
    c = getchar();
    if (c < '0' || c > '9') break;
    ret = 10 * ret + c - '0';
  }
  return sign * ret;
}
inline void ini(int& x) { x = getnum<int>(); }
inline void scani(int& x) { scanf("%d", &x); }
const int N = 1e5 + 5;
const int B = 17;
int n, m, q;
vector<int> lst[N];
int depth[N];
int sparset[B + 1][N];
vector<int> vec[B + 1][N];
void dfs(int now, int par) {
  sparset[0][now] = par;
  for (int i = 0; i < lst[now].size(); i++) {
    int to = lst[now][i];
    if (to == par) continue;
    depth[to] = depth[now] + 1;
    dfs(to, now);
  }
}
vector<int> combine(vector<int>& a, vector<int>& b) {
  vector<int> ret;
  int pa = 0, push_back = 0;
  while (pa < a.size() && push_back < b.size() && ret.size() < 10) {
    if (a[pa] < b[push_back])
      ret.push_back(a[pa++]);
    else
      ret.push_back(b[push_back++]);
  }
  while (pa < a.size() && ret.size() < 10) ret.push_back(a[pa++]);
  while (push_back < b.size() && ret.size() < 10) ret.push_back(b[push_back++]);
  return ret;
}
priority_queue<int> pq;
void add(vector<int>& v) {
  for (int i = 0; i < v.size(); i++) {
    if (pq.size() >= 10 && v[i] >= pq.top()) break;
    pq.push(v[i]);
    if (pq.size() > 10) pq.pop();
  }
}
void query(int a, int b) {
  if (depth[a] > depth[b]) swap(a, b);
  for (int i = B; i >= 0; i -= 1) {
    if (sparset[i][b] != -1 && depth[sparset[i][b]] >= depth[a]) {
      add(vec[i][b]);
      b = sparset[i][b];
    }
  }
  if (a == b) {
    add(vec[0][a]);
    return;
  }
  for (int i = B; i >= 0; i -= 1) {
    if (sparset[i][a] != sparset[i][b]) {
      add(vec[i][a]);
      add(vec[i][b]);
      a = sparset[i][a];
      b = sparset[i][b];
    }
  }
  add(vec[1][a]);
  add(vec[0][b]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.setf(ios::fixed);
  cout.setf(ios::showpoint);
  cout.precision(10);
  cin >> n >> m >> q;
  for (int i = 1; i <= n - 1; i += 1) {
    int a, b;
    cin >> a >> b;
    lst[a].push_back(b);
    lst[b].push_back(a);
  }
  depth[1] = 1;
  dfs(1, -1);
  for (int i = 1; i <= m; i += 1) {
    int x;
    cin >> x;
    vec[0][x].push_back(i);
  }
  for (int i = 1; i <= n; i += 1) {
    sort(vec[0][i].begin(), vec[0][i].end());
    if (vec[0][i].size() > 10) vec[0][i].resize(10);
  }
  for (int i = 0; i <= B - 1; i += 1) {
    for (int j = 1; j <= n; j += 1) {
      if (sparset[i][j] == -1) {
        sparset[i + 1][j] = -1;
        vec[i + 1][j] = vec[i][j];
      } else {
        sparset[i + 1][j] = sparset[i][sparset[i][j]];
        vec[i + 1][j] = combine(vec[i][j], vec[i][sparset[i][j]]);
      }
    }
  }
  while (q--) {
    int a, b, x;
    cin >> a >> b >> x;
    query(a, b);
    vector<int> tmp;
    while (!pq.empty()) {
      tmp.push_back(pq.top());
      pq.pop();
    }
    sort(tmp.begin(), tmp.end());
    cout << min((int)tmp.size(), x);
    for (int i = 0; i < min((int)tmp.size(), x); i++) {
      cout << " " << tmp[i];
    }
    cout << '\n';
  }
  return 0;
}